The point P `(3,2,4),Q(4,52),R(5,8,0) and S(2,-1,6)` are :

Options

A. Vertices of a rhombus which is not a square
B. Non coplanar
C. Collinear
D. Coplanar but not collinear

Answer

To determine the relationship between the points P(3, 2, 4), Q(4, 5, 2), R(5, 8, 0), and S(2, -1, 6), we will analyze their distances and positions in 3D space.

### Step-by-Step Solution:

1. **Identify the Points**:
- P(3, 2, 4)
- Q(4, 5, 2)
- R(5, 8, 0)
- S(2, -1, 6)

2. **Calculate the Distance PQ**:
- Use the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
- For PQ:
\[ PQ = \sqrt{(4 - 3)^2 + (5 - 2)^2 + (2 - 4)^2} \]
\[ = \sqrt{1^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \]

3. **Calculate the Distance QR**:
- For QR:
\[ QR = \sqrt{(5 - 4)^2 + (8 - 5)^2 + (0 - 2)^2} \]
\[ = \sqrt{1^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \]

4. **Calculate the Distance RS**:
- For RS:
\[ RS = \sqrt{(2 - 5)^2 + (-1 - 8)^2 + (6 - 0)^2} \]
\[ = \sqrt{(-3)^2 + (-9)^2 + 6^2} = \sqrt{9 + 81 + 36} = \sqrt{126} = 3\sqrt{14} \]

5. **Calculate the Distance PS**:
- For PS:
\[ PS = \sqrt{(2 - 3)^2 + (-1 - 2)^2 + (6 - 4)^2} \]
\[ = \sqrt{(-1)^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \]

6. **Analyze the Distances**:
- We found:
- PQ = √14
- QR = √14
- PS = √14
- RS = 3√14
- Since PQ, QR, and PS are equal and RS is different, we can conclude that the points are not forming a square or rhombus but they are not coplanar either.

7. **Conclusion**:
- The points are **non-coplanar** because they do not lie on the same plane. Therefore, the correct option is **B: non-coplanar**.